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The directional field for dydx=4xyin shown in figure 1.12.

(a) Verify that the straight lines y=±2xare solution curves, provided x≠0.

(b) Sketch the solution curve with initial condition y (0) = 2.

(c) Sketch the solution curve with initial condition y(2) = 1.

(d) What can you say about the behaviour of the above solution as x→+∞? How about x→-∞?

Short Answer

Expert verified
  1. Proved
  2. The graph is drawn below.
  3. The graph is drawn below.
  4. The curves are increasing and unbounded whenx→+∞orx→-∞.

Step by step solution

01

Step 1(a): Verify y=± 2x are solutions of a given curve.

As y=±2x

dydx=±24xy=±2          (dydx=4xy)4x±2x=±2       (x≠0)

This y=±2xis the solution curve of dydx=4xyfor any interval except x ≠ 0.

02

Step 2(b): Sketch the solution curve with initial condition y (0) = 2.

The curve is dydx=4xy

ydy=4xdx

Integrating on both sides

y22=4(x22)+cy22=2x2+c .....1

Put the point (0, 2) in equation (1)

(2)22=2(0)2+cc=2y22=2x2+2

Hence this is the solution curve with initial condition y (0) = 2.

03

Step 3(c): Sketch the solution curve with initial condition y(2) = 1.

The curve isy22=2x2+c.....2

Put the value of (2, 1) in equation (2)

(1)22=2(2)2+c8+c=12c=152y22=2x2+152

Hence this is the solution curve with initial condition y(2) = 1.

04

Step 4(d): Find the behaviour of the above solution

On solving parts (b) and (c), we get an increased unboundedly curve and have the lines

y = 2x andy=±2x as slant asymptotes asx→∞orx→-∞. And in part (c) also increases without bounds asx→∞and approaches to line y = 2x and not for x<0.

Therefore the curves are increasing and unbounded whenx→∞orx→-∞.

Hence, the solutions in parts (b) and (c) become infinite and have the line

y = 2x as an asymptote x→∞. As x→-∞, the solution in part (b) becomes infinite and has y = -2x as an asymptote, the solution in part (c) does not even exist for x negative.

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